Differential-Equation Solution

The differential-equation solution of the quantum harmonic oscillator derives both the discrete energies and the position-space eigenfunctions directly from square integrability. Its essential logic is:

1. scale out all dimensions;
2. determine the admissible large-distance behavior;
3. factor out the decaying Gaussian;
4. solve the remaining equation by a power series;
5. reject every series that regenerates exponential growth;
6. normalize the surviving Hermite-polynomial solutions.

The polynomial condition is therefore physical, not cosmetic. It is how the boundary condition at infinity selects isolated energies.

Stationary Equation

For

$$\hat H =-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +\frac12m\omega^2x^2,$$

the stationary Schrödinger equation is

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} +\frac12m\omega^2x^2\psi =E\psi.$$

An admissible bound-state solution must belong to $L^2(\mathbb R)$:

$$\int_{-\infty}^{\infty} \lvert\psi(x)\rvert^2\,dx<\infty.$$

The potential and its derivatives are finite everywhere, so $\psi$ and $d\psi/dx$ are continuous on the real line. There are no finite-position matching boundaries. The spectral information enters through decay at both infinities.

Dimensionless Form

Introduce the oscillator length and dimensionless coordinate

$$\ell=\sqrt{\frac{\hbar}{m\omega}}, \qquad \xi=\frac{x}{\ell},$$

and define

$$\epsilon=\frac{E}{\hbar\omega}.$$

Because

$$\frac{d}{dx} =\frac{1}{\ell}\frac{d}{d\xi}, \qquad \frac{\hbar^2}{m\ell^2} =\hbar\omega, \qquad m\omega^2\ell^2 =\hbar\omega,$$

the equation becomes

$$\frac12\left( -\frac{d^2}{d\xi^2}+\xi^2 \right)\psi(\xi) =\epsilon\psi(\xi).$$

Equivalently,

$$\frac{d^2\psi}{d\xi^2} +(2\epsilon-\xi^2)\psi=0.$$

All dependence on $m$, $\omega$, and $\hbar$ has disappeared. The dimensional parameters will re-enter only when the solution is rescaled to $x$ and $E$.

Some texts instead define a doubled spectral parameter $\lambda=2E/(\hbar\omega)$. Their equation is $-\psi''+\xi^2\psi=\lambda\psi$ and their final condition is $\lambda=2n+1$. This is the same convention written differently.

Large-Distance Behavior

For $\lvert\xi\rvert\gg\sqrt{2\epsilon}$, the $\xi^2$ term dominates:

$$\frac{d^2\psi}{d\xi^2} -\xi^2\psi\approx0.$$

To determine the leading exponential behavior, try

$$\psi(\xi)\sim e^{\alpha\xi^2/2}.$$

Then

$$\frac{d^2\psi}{d\xi^2} =\left(\alpha+\alpha^2\xi^2\right)\psi.$$

At leading order in $\xi^2$, the asymptotic equation requires

$$\alpha^2=1.$$

Thus the two asymptotic branches behave roughly as

$$\psi(\xi)\sim e^{-\xi^2/2} \quad\text{or}\quad \psi(\xi)\sim e^{+\xi^2/2}.$$

Only the first can be square-integrable. The omitted algebraic prefactors depend on the energy, but they cannot rescue the growing exponential. This analysis motivates extracting the decaying Gaussian before solving the remaining equation.

Factoring the Gaussian

Write

$$\psi(\xi)=h(\xi)e^{-\xi^2/2}.$$

The derivatives are

$$\frac{d\psi}{d\xi} =e^{-\xi^2/2} \left(h'-\xi h\right),$$

and

$$\frac{d^2\psi}{d\xi^2} =e^{-\xi^2/2} \left[ h''-2\xi h' +(\xi^2-1)h \right].$$

Substituting into $\psi''+(2\epsilon-\xi^2)\psi=0$ and dividing by the nonzero Gaussian gives

$$h''-2\xi h' +(2\epsilon-1)h=0.$$

This becomes the physicists’ Hermite equation when

$$2\epsilon-1=2n,$$

but that condition has not yet been established. It must follow from the behavior of the general solution.

Power-Series Recurrence

The differential equation has no finite singular points, so expand about the origin:

$$h(\xi)=\sum_{j=0}^{\infty}a_j\xi^j.$$

Then

$$h' =\sum_{j=0}^{\infty}(j+1)a_{j+1}\xi^j,$$

and

$$h'' =\sum_{j=0}^{\infty} (j+2)(j+1)a_{j+2}\xi^j.$$

After shifting indices in the $\xi h'$ term, the coefficient of each power $\xi^j$ must vanish:

$$(j+2)(j+1)a_{j+2} +(2\epsilon-1-2j)a_j=0.$$

Therefore

$$a_{j+2} =\frac{2j+1-2\epsilon} {(j+2)(j+1)}a_j.$$

The recurrence advances by two powers. It consequently generates two independent solutions:

• an even series fixed by $a_0$ with $a_1=0$;
• an odd series fixed by $a_1$ with $a_0=0$.

This is the differential-equation origin of definite parity. Since the one-dimensional oscillator has nondegenerate bound states, a physical eigenfunction cannot contain independent even and odd pieces at the same energy.

Why a Generic Series Fails

For large $j$ at fixed $\epsilon$, the recurrence behaves as

$$\frac{a_{j+2}}{a_j} \sim\frac{2}{j}.$$

The coefficients of $e^{\xi^2}$ have the same large-order ratio. Thus a nonterminating $h(\xi)$ generically develops the behavior

$$h(\xi)\sim e^{\xi^2}$$

in at least one asymptotic direction. Restoring the extracted Gaussian gives

$$\psi(\xi) =h(\xi)e^{-\xi^2/2} \sim e^{+\xi^2/2},$$

which is not square-integrable.

This argument should be interpreted with a little care. For an arbitrary trial energy, one can choose a solution that decays as $\xi\to+\infty$, but it then generally contains a growing component as $\xi\to-\infty$, or vice versa. A normalizable bound state must decay at both ends. Only isolated energies allow those two boundary requirements to be satisfied simultaneously.

Termination and Quantized Energy

The problematic growth disappears if the nonzero parity series terminates. Suppose its highest power is $\xi^n$ with $a_n\ne0$. The next coefficient vanishes only if

$$2n+1-2\epsilon=0.$$

Hence

$$\epsilon_n=n+\frac12, \qquad n=0,1,2,\ldots$$

and therefore

$$E_n=\hbar\omega\left(n+\frac12\right).$$

Once the numerator vanishes at $j=n$, all later coefficients of the same parity vanish. The surviving degree-$n$ polynomial is proportional to the physicists’ Hermite polynomial $H_n(\xi)$:

$$h_n(\xi)=C_nH_n(\xi).$$

For even $n$, the acceptable polynomial comes from the even series; for odd $n$, it comes from the odd series. The other parity solution at the same energy does not terminate and must be discarded.

Normalizability has now done two jobs at once: it has selected the discrete energies and the allowed parity branch at each energy.

First Polynomial Solutions

The Hermite equation for level $n$ is

$$H_n''-2\xi H_n'+2nH_n=0.$$

With the conventional leading coefficient $2^n$,

$$\begin{aligned} H_0(\xi)&=1,\\ H_1(\xi)&=2\xi,\\ H_2(\xi)&=4\xi^2-2,\\ H_3(\xi)&=8\xi^3-12\xi. \end{aligned}$$

The corresponding unnormalized eigenfunctions are

$$\begin{aligned} \psi_0(\xi)&\propto e^{-\xi^2/2},\\ \psi_1(\xi)&\propto \xi e^{-\xi^2/2},\\ \psi_2(\xi)&\propto (2\xi^2-1)e^{-\xi^2/2},\\ \psi_3(\xi)&\propto (2\xi^3-3\xi)e^{-\xi^2/2}. \end{aligned}$$

Multiplying a polynomial by a nonzero constant does not change the physical state after normalization, so the simplified polynomial factors above are equivalent to $H_n$.

Normalization

The physicists’ Hermite polynomials obey

$$\int_{-\infty}^{\infty} e^{-\xi^2}H_m(\xi)H_n(\xi)\,d\xi =\sqrt{\pi}\,2^n n!\,\delta_{mn}.$$

Let

$$\psi_n(x) =A_n H_n(x/\ell) e^{-x^2/(2\ell^2)}.$$

Using $dx=\ell\,d\xi$,

$$\begin{aligned} \delta_{mn} &=\int_{-\infty}^{\infty} \psi_m^*(x)\psi_n(x)\,dx\\ &=A_m^*A_n\ell \int_{-\infty}^{\infty} e^{-\xi^2}H_m(\xi)H_n(\xi)\,d\xi. \end{aligned}$$

Choosing real positive normalization constants gives

$$A_n =\frac{1} {\pi^{1/4}\sqrt{2^n n!\,\ell}}.$$

Thus

$$\psi_n(x) =\frac{1}{\sqrt{2^n n!}} \left(\frac{1}{\pi\ell^2}\right)^{1/4} H_n\left(\frac{x}{\ell}\right) e^{-x^2/(2\ell^2)}.$$

The factor $\ell^{-1/2}$ is required dimensionally: a normalized one-dimensional wavefunction has units of inverse square root of length.

The Hermite Functions page develops orthonormality, completeness, recurrences, and Fourier-transform identities without repeating this quantization derivation.

Parity, Nodes, and Nondegeneracy

Because

$$H_n(-\xi)=(-1)^nH_n(\xi),$$

the eigenfunctions satisfy

$$\psi_n(-x)=(-1)^n\psi_n(x).$$

$H_n$ has $n$ distinct real zeros, so $\psi_n$ has exactly $n$ nodes. This agrees with the Sturm oscillation theorem: in a one-dimensional confining potential, the $n$th bound state ordered from the bottom has $n$ nodes.

Nondegeneracy can be seen directly. If $\psi$ and $\chi$ are square-integrable solutions at the same energy, their Wronskian

$$W(x)=\psi(x)\chi'(x)-\psi'(x)\chi(x)$$

has derivative $W'(x)=0$. Both functions and their derivatives decay at infinity, so the constant Wronskian is zero. The two solutions are therefore linearly dependent. There is only one physical state, up to normalization and phase, at each $E_n$.

Direct Checks

For the ground state,

$$\psi_0(x)=A_0e^{-x^2/(2\ell^2)},$$

one finds

$$\frac{d^2\psi_0}{dx^2} =\left( \frac{x^2}{\ell^4}-\frac{1}{\ell^2} \right)\psi_0.$$

Substitution into the dimensional Schrödinger equation cancels the $x^2$ terms and leaves

$$\hat H\psi_0=\frac12\hbar\omega\,\psi_0.$$

For every $n$, three quick checks are available:

• parity must be $(-1)^n$;
• the polynomial degree and node count must be $n$;
• applying the Hamiltonian must produce $\hbar\omega(n+1/2)\psi_n$ with no residual power of $x$.

These checks often catch a missing factor of two in the Gaussian or confusion between the physicists’ and probabilists’ Hermite conventions.

Boundary Behavior and Forbidden Tails

The classical turning points are

$$x_{\mathrm t}^{(n)} =\pm\ell\sqrt{2n+1}.$$

They are not boundaries of the differential equation. The polynomial times Gaussian remains nonzero outside them and decays only as $\lvert x\rvert\to\infty$. This penetration is present even for the ground state, whose turning points lie at $\pm\ell$ while its Gaussian extends over all $x$.

The polynomial factor modifies the detailed tail but never defeats Gaussian decay:

$$\psi_n(x) \sim x^n e^{-x^2/(2\ell^2)} \quad \text{as }\lvert x\rvert\to\infty.$$

Any solution behaving like $e^{+x^2/(2\ell^2)}$ is excluded, regardless of its behavior over a finite plotting interval.

Shooting-Method Interpretation

The same quantization logic appears in numerical shooting. Parity supplies initial data at the origin:

$$\begin{array}{c|cc} \text{parity} & \psi(0) & \psi'(0)\\ \hline \text{even} & 1 & 0\\ \text{odd} & 0 & 1 \end{array}$$

for arbitrary overall normalization. Integrating outward at a trial energy almost always produces contamination by the exponentially growing branch. The eigenvalues are the isolated trial energies for which the coefficient of that branch vanishes.

Naive outward integration becomes ill-conditioned far into the forbidden region because even tiny numerical contamination eventually dominates the decaying solution. Stable alternatives include matching logarithmic derivatives from opposite sides, diagonalizing the Hamiltonian in a basis, or integrating only to a finite matching point while monitoring convergence.

Common Mistakes

• Defining $\epsilon=E/(\hbar\omega)$ but using formulas derived for $\lambda=2E/(\hbar\omega)$.
• Writing $e^{-x^2/\ell^2}$ rather than $e^{-x^2/(2\ell^2)}$ for the wavefunction.
• Treating the Gaussian factor as a lucky guess instead of deriving its asymptotic exponent.
• Saying the series terminates merely to make the answer simple.
• Keeping both even and odd series at one nondegenerate energy.
• Forgetting $dx=\ell\,d\xi$ during normalization.
• Mixing physicists’ $H_n(\xi)$ with probabilists’ Hermite polynomials.
• Mistaking the classical turning points for hard-wall boundary conditions.
• Assuming a numerical solution that looks bounded on a finite interval will remain normalizable at infinity.

Where This Is Used

• Quantum Harmonic Oscillator gives the physical overview and classical correspondence.
• Ladder-Operator Solution: First Encounter derives the same spectrum by factorization and positivity.
• Hermite Polynomials develops generating functions, recurrence relations, and orthogonality.
• Ordinary Differential Equations supplies the general theory of series solutions and boundary-value problems.
• Boundary Conditions places the decay requirement in the broader one-dimensional setting.
• Sturm–Liouville Theory explains orthogonality, ordering, and node counting.
• Number States translates the normalized functions into the energy-basis language.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018, sec. 2.3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, sec. 7.2.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, vol. 1, Wiley, 1977, complement V A.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Butterworth-Heinemann, 1977, sec. 23.
• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013, sec. 18.3.

Exercises

1. Verify every term in the dimensionless equation starting from the dimensional Schrödinger equation.

Solution

Using $x=\ell\xi$ and $\ell^2=\hbar/(m\omega)$,

$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} =-\frac{\hbar^2}{2m\ell^2}\frac{d^2}{d\xi^2} =-\frac{\hbar\omega}{2}\frac{d^2}{d\xi^2},$$

while

$$\frac12m\omega^2x^2 =\frac12m\omega^2\ell^2\xi^2 =\frac{\hbar\omega}{2}\xi^2.$$

Dividing the full equation by $\hbar\omega$ gives

$$\frac12\left( -\frac{d^2}{d\xi^2}+\xi^2 \right)\psi =\frac{E}{\hbar\omega}\psi.$$

2. Starting from $\psi=h e^{-\xi^2/2}$, derive the differential equation for $h$ without skipping the derivative terms.

Solution

Differentiation gives

$$\psi'=e^{-\xi^2/2}(h'-\xi h)$$

and

$$\psi'' =e^{-\xi^2/2} \left[h''-2\xi h'+(\xi^2-1)h\right].$$

Insert this into $\psi''+(2\epsilon-\xi^2)\psi=0$. The $\xi^2h$ terms cancel, leaving

$$h''-2\xi h'+(2\epsilon-1)h=0.$$

3. Use the recurrence relation to construct the polynomial for $n=2$ and identify its energy.

Solution

For termination at $n=2$,

$$\epsilon=2+\frac12=\frac52.$$

The even recurrence begins with arbitrary $a_0$:

$$a_2 =\frac{1-2\epsilon}{2}a_0 =-2a_0.$$

At $j=2$ the next numerator vanishes, so $a_4=0$. Thus

$$h(\xi)=a_0(1-2\xi^2).$$

This is proportional to $H_2(\xi)=4\xi^2-2$. The energy is $E_2=5\hbar\omega/2$.

4. Normalize the first excited state written as $\psi_1(x)=Cxe^{-x^2/(2\ell^2)}$.

Solution

Normalization requires

$$1 =\lvert C\rvert^2 \int_{-\infty}^{\infty} x^2e^{-x^2/\ell^2}\,dx.$$

The Gaussian moment is

$$\int_{-\infty}^{\infty} x^2e^{-x^2/\ell^2}\,dx =\frac{\sqrt\pi}{2}\ell^3.$$

Choosing $C$ real and positive gives

$$C=\frac{\sqrt2}{\pi^{1/4}\ell^{3/2}},$$

in agreement with the general formula.

5. Show from the Wronskian that two square-integrable oscillator solutions with the same energy must be proportional.

Solution

For two solutions $\psi$ and $\chi$ at the same $E$, subtract $\psi$ times the equation for $\chi$ from $\chi$ times the equation for $\psi$. The potential and energy terms cancel, leaving

$$\psi\chi''-\chi\psi''=0.$$

But

$$\frac{d}{dx} \left(\psi\chi'-\psi'\chi\right) =\psi\chi''-\psi''\chi,$$

so the Wronskian is constant. Bound-state solutions and their derivatives vanish at infinity, making that constant zero. A vanishing Wronskian for solutions of a second-order linear equation implies linear dependence.

6. In an even-parity shooting calculation, why can a tiny energy error be nearly invisible near the origin but catastrophic at large $\xi$?

Solution

Near the origin, both the physical and nonphysical solutions are finite, so a small admixture of the wrong solution can remain numerically small. In the forbidden region, however, the two independent asymptotic branches behave roughly as $e^{-\xi^2/2}$ and $e^{+\xi^2/2}$. Their ratio grows as $e^{\xi^2}$. Any nonzero coefficient of the growing branch eventually dominates, so outward shooting is exponentially sensitive to the trial energy and roundoff.